From 7dd5ce1f3d8213117d065b060a1609ba4536a19a Mon Sep 17 00:00:00 2001
From: goodyong <50802846+goodyong@users.noreply.github.com>
Date: Sat, 10 Jul 2021 12:20:02 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E4=BA=86python=E4=BB=A3?=
 =?UTF-8?q?=E7=A0=81=EF=BC=88=E6=95=B0=E7=BB=84=E7=AF=87=EF=BC=89?=
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为数组篇文件夹下的代码增加了python语言版本
---
 .../leetcode1438绝对值不超过限制的最长子数组.md | 35 ++++++++
 .../数组篇/leetcode1两数之和.md               | 35 ++++++++
 .../数组篇/leetcode219数组中重复元素2.md      | 53 ++++++++++++-
 .../数组篇/leetcode27移除元素.md              | 27 ++++++-
 .../数组篇/leetcode41缺失的第一个正数.md      | 23 ++++++
 .../数组篇/leetcode485最大连续1的个数.md      | 25 ++++++
 .../数组篇/leetcode54螺旋矩阵.md              | 37 +++++++++
 .../数组篇/leetcode560和为K的子数组.md        | 10 +++
 .../数组篇/leetcode59螺旋矩阵2.md             | 79 +++++++++++++++++++
 animation-simulation/数组篇/leetcode66加一.md | 20 +++++
 .../数组篇/leetcode75颜色分类.md              | 61 ++++++++++++++
 .../数组篇/剑指offer3数组中重复的数.md        | 35 ++++++++
 .../数组篇/长度最小的子数组.md                | 24 ++++++
 13 files changed, 461 insertions(+), 3 deletions(-)

diff --git a/animation-simulation/数组篇/leetcode1438绝对值不超过限制的最长子数组.md b/animation-simulation/数组篇/leetcode1438绝对值不超过限制的最长子数组.md
index a3ac8b1..000facd 100644
--- a/animation-simulation/数组篇/leetcode1438绝对值不超过限制的最长子数组.md
+++ b/animation-simulation/数组篇/leetcode1438绝对值不超过限制的最长子数组.md
@@ -43,6 +43,8 @@
 
 其实，我们只要把握两个重点即可，我们的 maxdeque 维护的是一个单调递减的双端队列，头部为当前窗口的最大值， mindeque 维护的是一个单调递增的双端队列，头部为窗口的最小值，即可。好啦我们一起看代码吧。
 
+Java Code:
+
 ```java
 class Solution {
     public int longestSubarray(int[] nums, int limit) {
@@ -76,3 +78,36 @@ class Solution {
 }
 ```
 
+Python Code:
+
+```python
+from typing import List
+import collections
+class Solution:
+    def longestSubarray(self, nums: List[int], limit: int)->int:
+        maxdeque = collections.deque()
+        mindeque = collections.deque()
+        leng = len(nums)
+        right = 0
+        left = 0
+        maxwin = 0
+        while right < leng:
+            while len(maxdeque) != 0 and maxdeque[-1] < nums[right]:
+                maxdeque.pop()
+            while len(mindeque) != 0 and mindeque[-1] > nums[right]:
+                mindeque.pop()
+            
+            maxdeque.append(nums[right])
+            mindeque.append(nums[right])
+            while (maxdeque[0] - mindeque[0]) > limit:
+                if maxdeque[0] == nums[left]:
+                    maxdeque.popleft()
+                if mindeque[0] == nums[left]:
+                    mindeque.popleft()
+                left += 1
+            # 保留最大窗口
+            maxwin = max(maxwin, right - left + 1)
+            right += 1
+        return maxwin
+```
+
diff --git a/animation-simulation/数组篇/leetcode1两数之和.md b/animation-simulation/数组篇/leetcode1两数之和.md
index 43be63b..89db733 100644
--- a/animation-simulation/数组篇/leetcode1两数之和.md
+++ b/animation-simulation/数组篇/leetcode1两数之和.md
@@ -33,6 +33,8 @@
 
 **题目代码**
 
+Java Code:
+
 ```java
 class Solution {
     public int[] twoSum(int[] nums, int target) {
@@ -55,6 +57,25 @@ class Solution {
 }
 ```
 
+Python3 Code:
+
+```python
+from typing import List
+class Solution:
+    def twoSum(nums: List[int], target: int)->List[int]:
+        if len(nums) < 2:
+            return [0]
+        rearr = [0] * 2
+        # 查询元素
+        for i in range(0, len(nums)):
+            for j in range(i + 1, len(nums)):
+                # 发现符合条件情况
+                if nums[i] + nums[j] == target:
+                    rearr[0] = i
+                    rearr[1] = j
+        return rearr
+```
+
 **哈希表**
 
 **解析**
@@ -122,5 +143,19 @@ const twoSum = function (nums, target) {
 };
 ```
 
+Python3 Code:
 
+```python 
+from typing import List
+class Solution:
+    def twoSum(self, nums: List[int], target: int)->List[int]:
+        m = {}
+        for i in range(0, len(nums)):
+            # 如果存在则返回
+            if (target - nums[i]) in m.keys():
+                return [m[target - nums[i]], i]
+            # 不存在则存入
+            m[nums[i]] = i
+        return [0]
+```
 
diff --git a/animation-simulation/数组篇/leetcode219数组中重复元素2.md b/animation-simulation/数组篇/leetcode219数组中重复元素2.md
index c167d1e..d5b5d9d 100644
--- a/animation-simulation/数组篇/leetcode219数组中重复元素2.md
+++ b/animation-simulation/数组篇/leetcode219数组中重复元素2.md
@@ -29,6 +29,8 @@
 
 这个题目和我们上面那个数组中的重复数字几乎相同，只不过是增加了一个判断相隔是否小于K位的条件，我们先用 HashMap 来做一哈，和刚才思路一致，我们直接看代码就能整懂
 
+Java Code:
+
 ```java
 class Solution {
     public boolean containsNearbyDuplicate(int[] nums, int k) {
@@ -41,9 +43,9 @@ class Solution {
         for (int i = 0; i < nums.length; i++) {
             // 如果含有
             if (map.containsKey(nums[i])) {
-                //判断是否小于K，如果小于则直接返回
+                //判断是否小于K，如果小于等于则直接返回
                 int abs = Math.abs(i - map.get(nums[i]));
-                if (abs <= k)  return true;//小于则返回        
+                if (abs <= k)  return true;//小于等于则返回        
             }
             //更新索引，此时有两种情况，不存在，或者存在时，将后出现的索引保存
             map.put(nums[i],i);
@@ -53,6 +55,29 @@ class Solution {
 }
 ```
 
+Python3 Code:
+
+```python
+from typing import List
+class Solution:
+    def containsNearbyDuplicate(self, nums: List[int], k: int)->bool:
+        # 特殊情况
+        if len(nums) == 0:
+            return False
+        # 字典
+        m = {}
+        for i in range(0, len(nums)):
+            # 如果含有
+            if nums[i] in m.keys():
+                # 判断是否小于K，如果小于等于则直接返回
+                a = abs(i - m[nums[i]])
+                if a <= k:
+                    return True# 小于等于则返回  
+            # 更新索引，此时有两种情况，不存在，或者存在时，将后出现的索引保存
+            m[nums[i]] = i
+        return False
+```
+
 **HashSet**
 
 **解析**
@@ -65,6 +90,8 @@ class Solution {
 
 **题目代码**
 
+Java Code
+
 ```java
 class Solution {
     public boolean containsNearbyDuplicate(int[] nums, int k) {
@@ -91,3 +118,25 @@ class Solution {
 }
 ```
 
+Python3 Code:
+
+```python
+from typing import List
+class Solution:
+    def containsNearbyDuplicate(self, nums: List[int], k: int)->bool:
+        # 特殊情况
+        if len(nums) == 0:
+            return False
+        # 集合
+        s = set()
+        for i in range(0, len(nums)):
+            # 如果含有，返回True
+            if nums[i] in s:
+                return True
+            # 添加新元素
+            s.add(nums[i])
+            # 维护窗口长度
+            if len(s) > k:
+                s.remove(nums[i - k])
+        return False
+```        
diff --git a/animation-simulation/数组篇/leetcode27移除元素.md b/animation-simulation/数组篇/leetcode27移除元素.md
index 1a14c5b..db107bc 100644
--- a/animation-simulation/数组篇/leetcode27移除元素.md
+++ b/animation-simulation/数组篇/leetcode27移除元素.md
@@ -50,6 +50,8 @@
 
 **题目代码**
 
+Java Code:
+
 ```java
 class Solution {
     public int removeElement(int[] nums, int val) {    
@@ -75,6 +77,29 @@ class Solution {
 }
 ```
 
+Python3 Code:
+
+```python
+from typing import List
+class Solution:
+    def removeElement(self, nums: List[int], val: int)->int:
+        # 获取数组长度
+        leng = len(nums)
+        if 0 == leng:
+            return 0
+        i = 0
+        while i < leng:
+            # 发现符合条件的情况
+            if nums[i] == val:
+                # 前移一位
+                for j in range(i, leng - 1):
+                    nums[j] = nums[j + 1]
+                i -= 1
+                leng -= 1
+            i += 1
+        return i
+```
+
 **双指针**
 
 快慢指针的做法比较有趣，只需要一个 for 循环即可解决，时间复杂度为 O(n) ,总体思路就是有两个指针，前面一个后面一个，前面的用于搜索需要删除的值，当遇到需要删除的值时，前指针直接跳过，后面的指针不动，当遇到正常值时，两个指针都进行移动，并修改慢指针的值。最后只需输出慢指针的索引即可。
@@ -100,7 +125,7 @@ class Solution {
                 if (nums[j] == val) {
                     continue;
                 }
-                // 不等于目标值时，则赋值给num[i],i++
+                // 不等于目标值时，则赋值给nums[i],i++
                 nums[i++] = nums[j];
           }
           return i;
diff --git a/animation-simulation/数组篇/leetcode41缺失的第一个正数.md b/animation-simulation/数组篇/leetcode41缺失的第一个正数.md
index 73dacaf..d22a7df 100644
--- a/animation-simulation/数组篇/leetcode41缺失的第一个正数.md
+++ b/animation-simulation/数组篇/leetcode41缺失的第一个正数.md
@@ -39,6 +39,8 @@
 
 ![缺失的第一个正数](https://cdn.jsdelivr.net/gh/tan45du/github.io.phonto2@master/myphoto/缺失的第一个正数.1it1cow5aa8w.gif)
 
+Java Code:
+
 ```java
 class Solution {
     public int firstMissingPositive(int[] nums) {
@@ -65,6 +67,27 @@ class Solution {
 }
 ```
 
+Python3 Code:
+
+```python
+from typing import List
+class Solution:
+    def firstMissingPositive(self, nums: List[int])->int:
+        if len(nums) == 0:
+            return 1
+        # 因为是返回第一个正整数，不包括 0，所以需要长度加1，细节1
+        res = [0] * (len(nums) + 1)
+        # 将数组元素添加到辅助数组中
+        for x in nums:
+            if x > 0 and x < len(res):
+                res[x] = x
+        # 遍历查找,发现不一样时直接返回
+        for i in range(1, len(res)):
+            if res[i] != i:
+                return i
+        # 缺少最后一个，例如 1，2，3此时缺少 4 ，细节2 
+        return len(res)
+```
 
 
 我们通过上面的例子了解这个解题思想，我们有没有办法不使用辅助数组完成呢？我们可以使用原地置换，直接在 nums 数组内，将值换到对应的索引处，与上个方法思路一致，只不过没有使用辅助数组，理解起来也稍微难理解一些。
diff --git a/animation-simulation/数组篇/leetcode485最大连续1的个数.md b/animation-simulation/数组篇/leetcode485最大连续1的个数.md
index 6157b54..dfdfa01 100644
--- a/animation-simulation/数组篇/leetcode485最大连续1的个数.md
+++ b/animation-simulation/数组篇/leetcode485最大连续1的个数.md
@@ -30,6 +30,8 @@
 
 下面我们直接看代码吧
 
+Java Code:
+
 ```java
 class Solution {
     public int findMaxConsecutiveOnes(int[] nums) {
@@ -58,6 +60,29 @@ class Solution {
 }
 ```
 
+Python3 Code:
+
+```python
+from typing import List
+class Solution:
+    def findMaxConsecutiveOnes(self, nums: List[int])->int:
+        leng = len(nums)
+        left = 0
+        right = 0
+        maxcount = 0
+        while right < leng:
+            if nums[right] == 1:
+                right += 1
+                continue
+            # 保存最大值
+            maxcount = max(maxcount, right - left)
+            # 跳过 0 的情况
+            while right < leng and nums[right] == 0:
+                right += 1
+            # 同一起点继续遍历
+            left = right
+        return max(maxcount, right - left)
+```
 
 
 刚才的效率虽然相对高一些，但是代码不够优美，欢迎各位改进，下面我们说一下另外一种情况，一个特别容易理解的方法。
diff --git a/animation-simulation/数组篇/leetcode54螺旋矩阵.md b/animation-simulation/数组篇/leetcode54螺旋矩阵.md
index 011e298..cb3c678 100644
--- a/animation-simulation/数组篇/leetcode54螺旋矩阵.md
+++ b/animation-simulation/数组篇/leetcode54螺旋矩阵.md
@@ -46,6 +46,8 @@
 
 题目代码：
 
+Java Code:
+
 ```java
 class Solution {
     public List<Integer> spiralOrder(int[][] matrix) {
@@ -83,5 +85,40 @@ class Solution {
 
 ```
 
+Python3 Code:
+
+```python
+from typing import List
+class Solution:
+    def spiralOrder(self, matrix: List[List[int]])->List[int]:
+        arr = []
+        left = 0
+        right = len(matrix[0]) - 1
+        top = 0
+        down = len(matrix) - 1
+        while True:
+            for i in range(left, right + 1):
+                arr.append(matrix[top][i])
+            top += 1
+            if top > down:
+                break
+            for i in range(top, down + 1):
+                arr.append(matrix[i][right])
+            right -= 1
+            if left > right:
+                break
+            for i in range(right, left - 1, -1):
+                arr.append(matrix[down][i])
+            down -= 1
+            if top > down:
+                break
+            for i in range(down, top - 1, -1):
+                arr.append(matrix[i][left])
+            left += 1
+            if left > right:
+                break
+        return arr
+```
+
 
 
diff --git a/animation-simulation/数组篇/leetcode560和为K的子数组.md b/animation-simulation/数组篇/leetcode560和为K的子数组.md
index 70402fe..8e81e3e 100644
--- a/animation-simulation/数组篇/leetcode560和为K的子数组.md
+++ b/animation-simulation/数组篇/leetcode560和为K的子数组.md
@@ -21,6 +21,8 @@
 
 这个题目的题意很容易理解，就是让我们返回和为 k 的子数组的个数，所以我们直接利用双重循环解决该题，这个是很容易想到的。我们直接看代码吧。
 
+Java Code:
+
 ```java
 class Solution {
     public int subarraySum(int[] nums, int k) {
@@ -41,6 +43,8 @@ class Solution {
 }
 ```
 
+Python3版本的代码会超时
+
 下面我们我们使用前缀和的方法来解决这个题目，那么我们先来了解一下前缀和是什么东西。其实这个思想我们很早就接触过了。见下图
 
 ![](https://cdn.jsdelivr.net/gh/tan45du/github.io.phonto2@master/myphoto/微信截图_20210113193831.4wk2b9zc8vm0.png)
@@ -61,6 +65,8 @@ presum [2] = presum[1] + nums[1],presum[3] = presum[2] + nums[2] ... 所以我
 
 直接上代码。
 
+Java Code:
+
 ```java
 class Solution {
     public int subarraySum(int[] nums, int k) {
@@ -86,6 +92,8 @@ class Solution {
 }
 ```
 
+Python3版本的代码也会超时
+
 我们通过上面的例子我们简单了解了前缀和思想，那么我们如果继续将其优化呢？
 
 **前缀和 + HashMap**
@@ -94,6 +102,8 @@ class Solution {
 
 其实我们在之前的两数之和中已经用到了这个方法，我们一起来回顾两数之和 HashMap 的代码.
 
+Java Code:
+
 ```java
 class Solution {
     public int[] twoSum(int[] nums, int target) {
diff --git a/animation-simulation/数组篇/leetcode59螺旋矩阵2.md b/animation-simulation/数组篇/leetcode59螺旋矩阵2.md
index 6824ea6..50cf94c 100644
--- a/animation-simulation/数组篇/leetcode59螺旋矩阵2.md
+++ b/animation-simulation/数组篇/leetcode59螺旋矩阵2.md
@@ -66,6 +66,8 @@
 
 题目代码：
 
+Java Code:
+
 ```java
 class Solution {
     public List<Integer> spiralOrder(int[][] matrix) {
@@ -103,10 +105,45 @@ class Solution {
 
 ```
 
+Python3 Code:
 
+```python
+from typing import List
+class Solution:
+    def spiralOrder(self, matrix: List[List[int]])->List[int]:
+        arr = []
+        left = 0
+        right = len(matrix[0]) - 1
+        top = 0
+        down = len(matrix) - 1
+        while True:
+            for i in range(left, right + 1):
+                arr.append(matrix[top][i])
+            top += 1
+            if top > down:
+                break
+            for i in range(top, down + 1):
+                arr.append(matrix[i][right])
+            right -= 1
+            if left > right:
+                break
+            for i in range(right, left - 1, -1):
+                arr.append(matrix[down][i])
+            down -= 1
+            if top > down:
+                break
+            for i in range(down, top - 1, -1):
+                arr.append(matrix[i][left])
+            left += 1
+            if left > right:
+                break
+        return arr
+```
 
 我们仅仅是将 54 反过来了，往螺旋矩阵里面插值，下面我们直接看代码吧,大家可以也可以对其改进，大家可以思考一下，如果修改能够让代码更简洁！
 
+Java Code:
+
 ```java
 class Solution {
     public int[][] generateMatrix(int n) {
@@ -147,3 +184,45 @@ class Solution {
 }
 ```
 
+Python3 Code:
+
+```python
+from typing import List
+import numpy as np
+class Solution:
+    def generateMatrix(self, n: int)->List[List[int]]:
+        arr = np.array([[0] * n] * n)
+        left = 0
+        right = n - 1
+        top = 0
+        buttom = n - 1
+        num = 1
+        numsize = n * n
+        while True:
+            for i in range(left, right + 1):
+                arr[top][i] = num
+                num += 1
+            top += 1
+            if num > numsize:
+                break
+            for i in range(top, buttom + 1):
+                arr[i][right] = num
+                num += 1
+            right -= 1
+            if num > numsize:
+                break
+            for i in range(right, left - 1, -1):
+                arr[buttom][i] = num
+                num += 1
+            buttom -= 1
+            if num > numsize:
+                break
+            for i in range(buttom, top - 1, -1):
+                arr[i][left] = num
+                num += 1
+            left += 1
+            if num > numsize:
+                break
+        return arr.tolist()
+```
+
diff --git a/animation-simulation/数组篇/leetcode66加一.md b/animation-simulation/数组篇/leetcode66加一.md
index e22be1b..3030fee 100644
--- a/animation-simulation/数组篇/leetcode66加一.md
+++ b/animation-simulation/数组篇/leetcode66加一.md
@@ -44,6 +44,8 @@
 
 我们可以根据当前位 余10来判断，这样我们就可以区分属于第几种情况了，大家直接看代码吧，很容易理解的。
 
+Java Code:
+
 ```java
 class Solution {
     public int[] plusOne(int[] digits) {
@@ -65,3 +67,21 @@ class Solution {
 }
 ```
 
+Python Code:
+
+```python
+from typing import List
+class Solution:
+    def plusOne(self, digits: List[int])->List[int]:
+        # 获取长度
+        leng = len(digits)
+        for i in range(leng - 1, -1, -1):
+            digits[i] = (digits[i] + 1) % 10
+            # 第一种和第二种情况，如果此时某一位不为 0 ，则直接返回即可。
+            if digits[i] != 0:
+                return digits
+        # 第三种情况，因为数组初始化每一位都为0，我们只需将首位设为1即可
+        arr = [0] * (leng + 1)
+        arr[0] = 1
+        return arr
+```
diff --git a/animation-simulation/数组篇/leetcode75颜色分类.md b/animation-simulation/数组篇/leetcode75颜色分类.md
index c14b9ae..7f125f1 100644
--- a/animation-simulation/数组篇/leetcode75颜色分类.md
+++ b/animation-simulation/数组篇/leetcode75颜色分类.md
@@ -38,6 +38,8 @@
 
 下面我们直接看代码吧，和三向切分基本一致。
 
+Java Code:
+
 ```java
 class Solution {
     public void sortColors(int[] nums) {
@@ -65,6 +67,35 @@ class Solution {
 }
 ```
 
+Python3 Code:
+
+```python
+from typing import List
+class Solution:
+    def sortColors(self, nums: List[int]): 
+        leng = len(nums)
+        left = 0
+        # 这里和三向切分不完全一致
+        i = left
+        right = leng - 1
+        while i <= right:
+            if nums[i] == 2:
+                self.swap(nums, i, right)
+                right -= 1
+            elif nums[i] == 0:
+                self.swap(nums, i, left)
+                i += 1
+                left += 1
+            else:
+                i += 1
+        return nums
+    
+    def swap(self, nums: List[int], i: int, j: int):
+        temp = nums[i]
+        nums[i] = nums[j]
+        nums[j] = temp
+```
+
 另外我们看这段代码，有什么问题呢？那就是我们即使完全符合时，仍会交换元素，这样会大大降低我们的效率。
 
 例如：[0,0,0,1,1,1,2,2,2]
@@ -83,6 +114,8 @@ class Solution {
 
 另一种代码表示
 
+Java Code:
+
 ```java
 class Solution {
     public void sortColors(int[] nums) {
@@ -114,3 +147,31 @@ class Solution {
 }
 ```
 
+Python3 Code:
+
+```python
+from typing import List
+class Solution:
+    def sortColors(self, nums: List[int]):
+        left = 0
+        leng = len(nums)
+        right = leng - 1
+        i = 0
+        while i <= right:
+            if nums[i] == 0:
+                self.swap(nums, i, left)
+                left += 1
+            if nums[i] == 2:
+                self.swap(nums, i, right)
+                right -= 1
+                # 如果不等于 1 则需要继续判断，所以不移动 i 指针，i--
+                if nums[i] != 1:
+                    i -= 1
+            i += 1    
+        return nums
+
+    def swap(self, nums: List[int], i: int, j: int):
+        temp = nums[i]
+        nums[i] = nums[j]
+        nums[j] = temp
+```
\ No newline at end of file
diff --git a/animation-simulation/数组篇/剑指offer3数组中重复的数.md b/animation-simulation/数组篇/剑指offer3数组中重复的数.md
index f817fdd..9d09871 100644
--- a/animation-simulation/数组篇/剑指offer3数组中重复的数.md
+++ b/animation-simulation/数组篇/剑指offer3数组中重复的数.md
@@ -45,6 +45,22 @@ class Solution {
 }
 ```
 
+Python Code:
+
+```python
+from typing import List
+class Solution:
+    def findRepeatNumber(self, nums: List[int])->int:
+        s = set()
+        for x in nums:
+            # 如果发现某元素存在，则返回
+            if x in s:
+                return x
+            # 存入集合
+            s.add(x)
+        return -1
+```
+
 #### **原地置换**
 
 **解析**
@@ -101,3 +117,22 @@ public:
 };
 ```
 
+Python3 Code:
+
+```python
+from typing import List
+class Solution:
+    def findRepeatNumber(self, nums: List[int])->int:
+        if len(nums) == 0:
+            return -1
+        for i in range(0, len(nums)):
+            while nums[i] != i:
+                # 发现重复元素
+                if nums[i] == nums[nums[i]]:
+                    return nums[i]
+                # 置换，将指针下的元素换到属于它的索引处
+                temp = nums[i]
+                nums[i] = nums[temp]
+                nums[temp] = temp
+        return -1
+```
diff --git a/animation-simulation/数组篇/长度最小的子数组.md b/animation-simulation/数组篇/长度最小的子数组.md
index 2e88fb0..a4dc825 100644
--- a/animation-simulation/数组篇/长度最小的子数组.md
+++ b/animation-simulation/数组篇/长度最小的子数组.md
@@ -82,3 +82,27 @@ public:
 };
 ```
 
+Python3 Code:
+
+```python 
+from typing import List
+import sys
+class Solution:
+    def minSubArrayLen(self, s: int, nums: List[int])->int:
+        leng = len(nums)
+        windowlen = sys.maxsize
+        i = 0
+        sum = 0
+        for j in range(0, leng):
+            sum += nums[j]
+            while sum >= s:
+                windowlen = min(windowlen, j - i + 1)
+                sum -= nums[i]
+                i += 1
+            
+        if windowlen == sys.maxsize:
+            return 0
+        else:
+            return windowlen
+```
+